Both notations are right. The first one is already simplified, whereas the second one can be simplified further by explicitely computing the exterior product and you will end up with the first version. So I actually think the first version is cleaner.

The differential form in your integral is an assignment of an element of [itex]\Lambda(T^*_x U)[/itex] to every point [itex]x[/itex] of [itex]U[/itex]. [itex]\Lambda(T^*_x U)[/itex] is a vector space and [itex]f(x)[/itex] is just a coefficient (which is different at every point).

Both notations are right. The first one is already simplified, whereas the second one can be simplified further by explicitely computing the exterior product and you will end up with the first version. So I actually think the first version is cleaner.

The differential form in your integral is an assignment of an element of [itex]\Lambda(T^*_x U)[/itex] to every point [itex]x[/itex] of [itex]U[/itex]. [itex]\Lambda(T^*_x U)[/itex] is a vector space and [itex]f(x)[/itex] is just a coefficient (which is different at every point).

I'm sorry, I should have specified what I mean.

I'm talking mainly about the fact that ##f(x)## is the output of ##f##, whereas ##f## is the function.

I'm talking mainly about the fact that ##f(x)## is the output of ##f##, whereas ##f## is the function.

In that case I would write [itex]f \,\mathrm d x_1\wedge\ldots\wedge \mathrm d x_n[/itex], however, omitting the first wedge. If you know that [itex]f[/itex] is a 0-form, then you can simplify the exterior product. But you are right: Inserting the coordinates into f gives you the form at a point already, whereas the form itself should be written without inserting [itex]x[/itex] into [itex]f[/itex].